# The Adiabatic Theorem and Symmetries of the Hamiltonian

For this question, all operators and states are on a finite dimensional Hilbert space.

Suppose I have a collection of continuously parametrized Hamiltonians $$H(t), 0\leq t\leq T$$. Suppose furthermore that I have a time-independent Hermitian operator $$O$$ such that $$[H(t), O] = 0$$ for all $$t$$. Informally, the adiabatic theorem states that if $$|\psi(0)\rangle$$ is an eigenstate of $$H(0)$$, then, provided the evolution of $$H(t)$$ is sufficiently slow/long, $$|\psi(t)\rangle$$ will remain an eigenstate of $$H(t)$$ for all $$t$$. I am wondering what I can say about the relationship between $$|\psi(t)\rangle$$ and the symmetry operator $$O$$. Suppose for instance that $$|\psi(0)\rangle$$ is also an eigenstate of $$O$$ with eigenvalue $$\lambda$$. My questions are

1. Is it true that $$|\psi(t)\rangle$$ is an eigenstate of $$O$$ for all $$t$$? If so, will it necessarily have the same eigenvalue $$\lambda$$ as $$|\psi(0)\rangle$$?
2. Does the physics change if the eigenspace of $$O$$ associated to $$\lambda$$ is degenerate?

1. If $$| \psi(0) \rangle$$ is initially an eigenvalue of $$O$$ with eigenvalue $$\lambda$$, and $$O$$ commutes with your Hamiltonian $$H(t)$$ for all times $$t$$, then indeed the state will always be an eigenvalue of $$O$$ with eigenvalue $$\lambda$$. This follows trivially by writing the time evolution operator in terms of $$H$$. The most general form possible for the time evolution operator is obtained by the time ordered exponential $$U(t, t_0) = T \exp \Bigg\{ -i \int_{t_0}^t dt' H(t') \Bigg\} = \sum_{n = 0}^{\infty} \frac{(-i)^n}{n!} \int_{t_0}^t dt_1 \ldots \int_{t_0}^t dt_n \ T \Big\{ H(t_1) \ldots H(t_n) \Big\},$$ where $$T$$ is the time-ordering symbol, which orders the $$n$$ instances of the Hamiltonian from latest to earliest in time. This form for the time evolution operator is necessary when (for instance) $$H$$ does not commute with itself at different times. In any case, you can immediately see that $$[O,H(t)] = 0$$ implies $$[O,U(t)] = 0$$ (I've set $$t_0 = 0$$), and therefore $$O |\psi(t) \rangle = O U(t) |\psi(0) \rangle = U(t) O | \psi(0) \rangle = \lambda U(t) | \psi(0) \rangle = \lambda | \psi(t) \rangle$$
2. From the above, if the eigenspace associated to $$\lambda$$ is nondegenerate then clearly the state $$| \psi(t) \rangle$$ is proportional to $$| \psi(0) \rangle$$ at all times. On the other hand, if the eigenspace is degenerate then there's no reason $$| \psi(t) \rangle$$ can't explore the full eigenspace. As a trivial example, take $$O = 1$$ to be the identity operator on a 2-dimensional Hilbert space, and let $$H(t)$$ be anything that evolves nontrivially.